Question

A four-lane freeway (two lanes in each direction) operates at capacity during the peak hour. It has 11-ft lanes, 4-ft shoulders, and there are 3 ramps within 3 miles upstream of the segment midpoint and 4 ramps within 3 miles downstream of the segment midpoint. The freeway is on rolling terrain and has 8% heavy vehicles with a peak-hour factor of 0.85. It is know that 12% of the AADT occurs in the peak hour and that the directional factor is 0.6. What is the freeway’s AADT?

Answer 1

Assume Base free flow speed (BFFS) = 75 mph

Lane width = 11 ft

Reduction in speed corresponding to lane width, f_LW = 1.9 mph

Lateral Clearance = 4 ft

Reduction in speed corresponding to lateral clearance, f_LC = 1.2 mph

Interchanges/Ramps = 7 / 6 miles = 1.17 / mile

Reduction in speed corresponding to Interchanges/ramps, f_ID = 3.7 mph

No. of lanes = 2

Reduction in speed corresponding to number of lanes, f_N = 4.5 mph

Free Flow Speed (FFS) = BFFS – f_LW – f_LC – f_N – f_ID= 75 – 1.9 – 1.2 – 4.5 – 3.7 = 63.7 mph

Peak-hour factor = 0.85

Trucks = 8 %

Rolling Terrain

f_HV = 1/ (1 + 0.08 (2.5-1)) = 1/1.12 = 0.893

f_P = 1.0

Peak Flow Rate, V_p = V / (PHV*n*f_HV*f_P) = V/ (0.85*2*0.893*1.0) = V/1.5181 veh/hr/ln

S = FFS

S = 63.7 mph

Density = V_p/S

Assume Level of Service is E

Density of LOS E should lie between 36 – 45 veh/mi/ln

Assume density = 36 veh/mi/ln

Peak flow rate = V/1.5181 = 36 * 63.7

V = 3481.3 veh/hr ~ 3482 veh/hr

V = AADT * K * D = AADT * 0.12 * 0.6

AADT = 3482/(0.12 * 0.6) = 48361 veh/day

Freeway AADT = 48361 veh/day